Solution — 2025 Question 13 Binomial Distribution| Sampling

(a)

\begin{matrix} Unbiased estimate of population mean \\ = \overline{x} \\ = \frac{\sum (x - 60)}{n} + 60 \\ = \frac{72}{50} + 60 \\ = \frac{1536}{25} \\ = 61.44 ∎ \end{matrix}

\begin{matrix} Unbiased estimate of population variance \\ = s^{2} \\ = \frac{1}{n - 1} (\sum (x - 60)^{2} - \frac{\sum (x - 60)^{2}}{n}) \\ = \frac{1}{49} (758 - \frac{{(72)}^{2}}{50}) \\ = \frac{16 358}{1225} \\ = 13.4 ∎ (to 3 s.f.) \end{matrix}

(b)

Let $\overline{X}$ be the mean mass (in grams) of a randomly chosen tray of $36$ eggs.

By the Central Limit Theorem, since $n = 36$ is large,

$\overline{X} \sim N (61.44, \frac{13.353}{36}) approximately$

\begin{matrix} P (61 < \overline{X} < 63) & = 0.759 78 \\ = 0.760 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(61 < \bar{X} < 63\right)

(c)

Let $T$ represent the number of trays (out of 10) where the mean mass of an egg lies between 61 g and 63 g.

$T \sim B (10, 0.759 78)$

\begin{matrix} P (T \geq 7) & = 1 - P (T \leq 6) \\ = 1 - 0.201 75 \\ = 0.798 25 \\ = 0.798 ∎ (to 3 s.f.) \end{matrix}

probability of T \geq 7